Roger's Equations

A Hypersphere is an object that satisifies the following equation;

where n is the number of dimensions.

Hyperspheres of 1, 2, and 3 dimensions (n=1,2,3) are:

For n=1, the equation above looks like;

x₁= ±R

Which are two points, one located at -R and the other located at R

For n=2, the above equation looks like;

x₁² + x₂² = R²

Which is the equation of a circle with radius R.

For n=3, the equation looks like;

x₁² + x₂² + x₃² = R²

Which is the equation of a sphere

So a one dimension Hypersphere is two points at R and -R, a two dimension hypersphere is a circle, and a three dimension hypersphere is a sphere. For dimensions larger than three (4,5,...,n), the hypersphere is simply called a 4-Hypersphere, 5-Hypersphere, .... n-Hypersphere.

Notice in the equation above that each new dimension adds a new variable(x₁,x₂,x₃,...x_n), but the equation for a hypersphere, regardless of dimension, pretty much says that every point on the surface of the n-Hypersphere is R distance from its center.

So the equation for a 4 dimesional hypersphere is;

x₁² + x₂² + x₃² + x₄² = R²

I don't think we gain much by simplifying here. At this point we have an idea what a hypersphere is. Now lets get some properties of hyperspheres so we can understand them better.


Calculating the Volume and Surface Area of an n-Hypersphere

To find the volume of a three dimesional cube, we can simply solve the integral;

where f(x,y,z)=1 and x, y , and z range from 0 to D, where D is the length of a side of the cube. The result of the integral is;

D x D x D = Volume = D³

Similiarly, we can find the volume of a cuboid by again setting f(x,y,z)=1 and having x integrate from 0 to L, y integrate from 0 to W, and z integrate from 0 to H. The result of the integral is;

L x W x H = Volume

If we want to find the volume of a Sphere, we should switch to spherical coordinates since it will make soving the integral easier. We can use the following equations to switch to spherical coordinates;

x = r sinθ cosφ
y = r sinθ sinφ
z = r cosθ

If we want to do integrals in spherical coordinates, we need the volume element (dV = dxdydz). In spherical coordinates, instead of dxdydz, we need |J(r,θ,φ)|drdθdφ.

|J(r,θ,φ)| is known as the Jacobian and can be found by solving a specific determinant. In the two dimensional case, the determinent is:

J(r,) = = = r(cos² + sin²) = r.

So da= dxdy = rdrdθ

For three dimensions, dV = dxdydz = r²sinθdrdφdθ

So our integral in spherical coordinates is:

Volume= ∫∫∫ r²sinθ drdθdφ

We've replaced dxdydz with r²sinθ drdφdθ. we integrate r from 0 to R, φ from 0 to 2π, and θ from 0 to π. Integrating we get;

Volume = (R³/3) x 2π x (1 - (-1)) = (⁴/₃)πR³

So the volume of a sphere is (⁴/₃)πR³

If we want to find the surface area of the sphere, we should integrate only the angular parts:

Surface Area = r² ∫∫ sinθ drdθdφ

So integrating we get:

Surface Area = r² x (2π) x (2) = 4πr²

Ok, so now we know we can calculate the volume and surface area of a sphere in 3 dimensions (not to mention a cube or cuboid). How about a 4-hypersphere in four dimensions?


4 Dimensional Hypersphere

The trick here is to not try and visualize what your trying to measure. You're much better off trusting the math, since we know that it works in three and two (check for yourself) dimesions. It should be valid to extrapolate for 4 dimensions using similiar methods as above, namely converting to spherical coordinates and integrating.

In cartesian coordinates we know from above that 4 dimensional cartesian space would have the following variables;

x₁, x₂, x₃, and x₄

In 4 dimensional spherical coordinates we use the variables

r, θ₁, θ₂, and θ₃

Again we need relations between these coordinates, they are;

x = r sinθ₃ sinθ₂ cosθ₁
y = r sinθ₃ sinθ₂ sinθ₁
z = r sinθ₃ cosθ₂
x4= r cosθ₃

Now we can use the above relation to determine the Jacobian for 4 dimensional spherical coordinates

|J| = r³Sin²θ₃Sinθ₂

which tells us that the volume element dV = dx₁dx₂dx₃dx₄ = r³sin²θ₃sinθ₂drdθ₁dθ₂dθ₃
gives us the integral;

∫∫∫∫r³sin²θ₃sinθ₂drdθ₁dθ₂dθ₃

where integration for r is from 0 to R, for θ₁ is from 0 to 2π, for θ₂ is from 0 to π, and from θ₃ is from 0 to π. Integrating we get;

Volume = (R⁴/4) x (2π) x (2) x ((¹/₂)π) = (1/2)π²R⁴

We can find the surface area of a 4 dimensional Hypersphere by integrating over the angular parts of the volume integral, just like we did for the three dimensional sphere.

Surface Area =R³ ∫∫∫sin²θ₃sinθ₂dθ₁dθ₂dθ₃

Surface Area = R³ x (2π) x (2) x ((¹/₂)π) = 2π²R³

so from above we know that a 4 Dimesional Hypersphere of radius (R) = 1 has:

Surface Area = 2π²

and

Volume = (1/2)π²


Conclusions

So we can find the volume and surface area of a n-Hypersphere of 1, 2, 3, and 4 dimensions. In part II, we will develop a general expression for the volume and surface area for a n-Hypersphere with any dimension (n). This will allow us to see how volume and surface area changes with increasing number of dimensions, which will provide some surprising results.

Till next time.

Special thanks to the following websites:

http://en.wikipedia.org/wiki/Hypersphere
http://mathworld.wolfram.com/Hypersphere.html